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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #1211 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-cos{(x^3)}sin{(z^3)}\\&\text{In the direction of }\overrightarrow{v}=(0,8,-10).\end{align*}$

Possible Answers:

${2.34z^2cos{(x^3)}cos{(z^3)}}$

${38.4x^2sin{(x^3)}sin{(z^3)} - 38.4z^2cos{(x^3)}cos{(z^3)}}$

${-1.83z^2cos{(x^3)}cos{(z^3)}}$

Correct answer:

${2.34z^2cos{(x^3)}cos{(z^3)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(8)^2+(-10)^2}=12.81\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{12.81}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{8}{12.81}=0.62\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-10}{12.81}=-0.78\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(3x^2sin{(x^3)}sin{(z^3)})+(0.62)(0.0)+(-0.78)(-3z^2cos{(x^3)}cos{(z^3)})\\&D_{\overrightarrow{u}}(x,y,z)=2.34z^2cos{(x^3)}cos{(z^3)}\end{align*}$

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Example Question #1211 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{{(cos{(z^3)}sin{(x)})}}{y^2}\\&\text{In the direction of }\overrightarrow{v}=(8,9,0).\end{align*}$

Possible Answers:

${\frac{{(0.66cos{(z^3)}cos{(x)})}}{y^2} -\frac{ {(1.5cos{(z^3)}sin{(x)})}}{y^3}}$

${\frac{{(12cos{(z^3)}cos{(x)})}}{y^2} -\frac{ {(24.1cos{(z^3)}sin{(x)})}}{y^3} -\frac{ {(36.1z^2sin{(z^3)}sin{(x)})}}{y^2}}$

${\frac{{(8cos{(z^3)}cos{(x)})}}{y^2} -\frac{ {(18cos{(z^3)}sin{(x)})}}{y^3}}$

${\frac{{(72.2z^2sin{(z^3)}cos{(x)})}}{y^3}}$

Correct answer:

${\frac{{(0.66cos{(z^3)}cos{(x)})}}{y^2} -\frac{ {(1.5cos{(z^3)}sin{(x)})}}{y^3}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(8)^2+(9)^2+(0)^2}=12.04\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{8}{12.04}=0.66\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{9}{12.04}=0.75\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{12.04}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.66)(\frac{{(cos{(z^3)}cos{(x)})}}{y^2})+(0.75)(\frac{-{(2cos{(z^3)}sin{(x)})}}{y^3})+(0)(\frac{-{(3z^2sin{(z^3)}sin{(x)})}}{y^2})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(0.66cos{(z^3)}cos{(x)})}}{y^2} -\frac{ {(1.5cos{(z^3)}sin{(x)})}}{y^3}\end{align*}$

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Example Question #1211 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^3cos{(z^2)}ln{(x^2)}\\&\text{In the direction of }\overrightarrow{v}=(-4,0,-14).\end{align*}$

Possible Answers:

${\frac{-{(175y^2zsin{(z^2)})}}{x}}$

${\frac{{(29.1y^3cos{(z^2)})}}{x} + 43.7y^2cos{(z^2)}ln{(x^2)} - 29.1y^3zln{(x^2)}sin{(z^2)}}$

${1.92y^3zln{(x^2)}sin{(z^2)} -\frac{ {(0.54y^3cos{(z^2)})}}{x}}$

${28y^3zln{(x^2)}sin{(z^2)} -\frac{ {(8y^3cos{(z^2)})}}{x}}$

Correct answer:

${1.92y^3zln{(x^2)}sin{(z^2)} -\frac{ {(0.54y^3cos{(z^2)})}}{x}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(0)^2+(-14)^2}=14.56\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{14.56}=-0.27\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{14.56}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-14}{14.56}=-0.96\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.27)(\frac{{(2y^3cos{(z^2)})}}{x})+(0)(3y^2cos{(z^2)}ln{(x^2)})+(-0.96)(-2y^3zln{(x^2)}sin{(z^2)})\\&D_{\overrightarrow{u}}(x,y,z)=1.92y^3zln{(x^2)}sin{(z^2)} -\frac{ {(0.54y^3cos{(z^2)})}}{x}\end{align*}$

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Example Question #1212 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{{(cos{(x^2)}cos{(z)})}}{y^2}\\&\text{In the direction of }\overrightarrow{v}=(0,6,-11).\end{align*}$

Possible Answers:

${\frac{-{(50.1xsin{(x^2)}sin{(z)})}}{y^3}}$

${-\frac{ {(25.1cos{(x^2)}cos{(z)})}}{y^3} -\frac{ {(12.5cos{(x^2)}sin{(z)})}}{y^2} -\frac{ {(25.1xsin{(x^2)}cos{(z)})}}{y^2}}$

${-\frac{ {(0.461cos{(x^2)}cos{(z)})}}{y^3} -\frac{ {(0.774cos{(x^2)}sin{(z)})}}{y^2}}$

${\frac{{(0.88cos{(x^2)}sin{(z)})}}{y^2} -\frac{ {(0.96cos{(x^2)}cos{(z)})}}{y^3}}$

Correct answer:

${\frac{{(0.88cos{(x^2)}sin{(z)})}}{y^2} -\frac{ {(0.96cos{(x^2)}cos{(z)})}}{y^3}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(6)^2+(-11)^2}=12.53\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{12.53}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{6}{12.53}=0.48\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-11}{12.53}=-0.88\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{-{(2xsin{(x^2)}cos{(z)})}}{y^2})+(0.48)(\frac{-{(2cos{(x^2)}cos{(z)})}}{y^3})+(-0.88)(\frac{-{(1cos{(x^2)}sin{(z)})}}{y^2})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(0.88cos{(x^2)}sin{(z)})}}{y^2} -\frac{ {(0.96cos{(x^2)}cos{(z)})}}{y^3}\end{align*}$

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Example Question #1213 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^{(y^2)}x^3z^2\\&\text{In the direction of }\overrightarrow{v}=(15,17,0).\end{align*}$

Possible Answers:

${45\cdot3^{(y^2)}x^2z^2 + 37.4\cdot3^{(y^2)}x^3yz^2}$

${299\cdot3^{(y^2)}x^2yz}$

${1.98\cdot3^{(y^2)}x^2z^2 + 1.65\cdot3^{(y^2)}x^3yz^2}$

${45.3\cdot3^{(y^2)}x^3z + 68\cdot3^{(y^2)}x^2z^2 + 49.8\cdot3^{(y^2)}x^3yz^2}$

Correct answer:

${1.98\cdot3^{(y^2)}x^2z^2 + 1.65\cdot3^{(y^2)}x^3yz^2}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(15)^2+(17)^2+(0)^2}=22.67\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{15}{22.67}=0.66\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{17}{22.67}=0.75\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{22.67}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.66)(3\cdot3^{(y^2)}x^2z^2)+(0.75)(2.2\cdot3^{(y^2)}x^3yz^2)+(0)(2\cdot3^{(y^2)}x^3z)\\&D_{\overrightarrow{u}}(x,y,z)=1.98\cdot3^{(y^2)}x^2z^2 + 1.65\cdot3^{(y^2)}x^3yz^2\end{align*}$

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Example Question #1214 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{{(z^4ln{(x^2)})}}{y}\\&\text{In the direction of }\overrightarrow{v}=(0,15,12).\end{align*}$

Possible Answers:

${\frac{{(48z^3ln{(x^2)})}}{y} -\frac{ {(15z^4ln{(x^2)})}}{y^2}}$

${\frac{-{(154z^3)}}{{(xy^2)}}}$

${\frac{{(2.48z^3ln{(x^2)})}}{y} -\frac{ {(0.78z^4ln{(x^2)})}}{y^2}}$

${\frac{{(76.8z^3ln{(x^2)})}}{y} -\frac{ {(19.2z^4ln{(x^2)})}}{y^2} +\frac{ {(38.4z^4)}}{{(xy)}}}$

Correct answer:

${\frac{{(2.48z^3ln{(x^2)})}}{y} -\frac{ {(0.78z^4ln{(x^2)})}}{y^2}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(15)^2+(12)^2}=19.21\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{19.21}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{15}{19.21}=0.78\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{12}{19.21}=0.62\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{{(2z^4)}}{{(xy)}})+(0.78)(\frac{-{(1z^4ln{(x^2)})}}{y^2})+(0.62)(\frac{{(4z^3ln{(x^2)})}}{y})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(2.48z^3ln{(x^2)})}}{y} -\frac{ {(0.78z^4ln{(x^2)})}}{y^2}\end{align*}$

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Example Question #1215 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^3ln{(z^2)}sin{(y^3)}\\&\text{In the direction of }\overrightarrow{v}=(6,-10,0).\end{align*}$

Possible Answers:

${0.78x^2ln{(z^2)}sin{(y^3)} + 2.22x^3y^2cos{(y^3)}ln{(z^2)}}$

${\frac{{(23.3x^3sin{(y^3)})}}{z} + 35x^2ln{(z^2)}sin{(y^3)} + 35x^3y^2cos{(y^3)}ln{(z^2)}}$

${1.53x^2ln{(z^2)}sin{(y^3)} - 2.58x^3y^2cos{(y^3)}ln{(z^2)}}$

${\frac{{(210x^2y^2cos{(y^3)})}}{z}}$

Correct answer:

${1.53x^2ln{(z^2)}sin{(y^3)} - 2.58x^3y^2cos{(y^3)}ln{(z^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(6)^2+(-10)^2+(0)^2}=11.66\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{6}{11.66}=0.51\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-10}{11.66}=-0.86\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{11.66}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.51)(3x^2ln{(z^2)}sin{(y^3)})+(-0.86)(3x^3y^2cos{(y^3)}ln{(z^2)})+(0)(\frac{{(2x^3sin{(y^3)})}}{z})\\&D_{\overrightarrow{u}}(x,y,z)=1.53x^2ln{(z^2)}sin{(y^3)} - 2.58x^3y^2cos{(y^3)}ln{(z^2)}\end{align*}$

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Example Question #1212 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-x^3ze^{(y)}\\&\text{In the direction of }\overrightarrow{v}=(-7,0,7).\end{align*}$

Possible Answers:

${- 9.9x^3e^{(y)} - 29.7x^2ze^{(y)} - 9.9x^3ze^{(y)}}$

${-29.7x^2e^{(y)}}$

${21x^2ze^{(y)} - 7x^3e^{(y)}}$

${2.13x^2ze^{(y)} - 0.71x^3e^{(y)}}$

Correct answer:

${2.13x^2ze^{(y)} - 0.71x^3e^{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-7)^2+(0)^2+(7)^2}=9.9\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-7}{9.9}=-0.71\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{9.9}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{7}{9.9}=0.71\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.71)(-3x^2ze^{(y)})+(0)(-1x^3ze^{(y)})+(0.71)(-1x^3e^{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=2.13x^2ze^{(y)} - 0.71x^3e^{(y)}\end{align*}$

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Example Question #152 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{cos{(x)}}{{(y^2z)}}\\&\text{In the direction of }\overrightarrow{v}=(0,9,15).\end{align*}$

Possible Answers:

${-\frac{ {(17.5cos{(x)})}}{{(y^2z^2)}} -\frac{ {(35cos{(x)})}}{{(y^3z)}} -\frac{ {(17.5sin{(x)})}}{{(y^2z)}}}$

${-\frac{ {(0.74cos{(x)})}}{{(y^2z^2)}} -\frac{ {(0.52cos{(x)})}}{{(y^3z)}}}$

${-\frac{ {(0.86cos{(x)})}}{{(y^2z^2)}} -\frac{ {(1.02cos{(x)})}}{{(y^3z)}}}$

${\frac{-{(35sin{(x)})}}{{(y^3z^2)}}}$

Correct answer:

${-\frac{ {(0.86cos{(x)})}}{{(y^2z^2)}} -\frac{ {(1.02cos{(x)})}}{{(y^3z)}}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(9)^2+(15)^2}=17.49\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{17.49}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{9}{17.49}=0.51\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{15}{17.49}=0.86\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{-{(1sin{(x)})}}{{(y^2z)}})+(0.51)(\frac{-{(2cos{(x)})}}{{(y^3z)}})+(0.86)(\frac{-{(1cos{(x)})}}{{(y^2z^2)}})\\&D_{\overrightarrow{u}}(x,y,z)=-\frac{ {(0.86cos{(x)})}}{{(y^2z^2)}} -\frac{ {(1.02cos{(x)})}}{{(y^3z)}}\end{align*}$

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Example Question #153 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^3z^3sin{(x)}\\&\text{In the direction of }\overrightarrow{v}=(20,12,0).\end{align*}$

Possible Answers:

${210y^2z^2cos{(x)}}$

${23.3y^3z^3cos{(x)} + 70y^2z^3sin{(x)} + 70y^3z^2sin{(x)}}$

${20y^3z^3cos{(x)} + 36y^2z^3sin{(x)}}$

${0.86y^3z^3cos{(x)} + 1.53y^2z^3sin{(x)}}$

Correct answer:

${0.86y^3z^3cos{(x)} + 1.53y^2z^3sin{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(20)^2+(12)^2+(0)^2}=23.32\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{20}{23.32}=0.86\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{12}{23.32}=0.51\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{23.32}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.86)(y^3z^3cos{(x)})+(0.51)(3y^2z^3sin{(x)})+(0)(3y^3z^2sin{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=0.86y^3z^3cos{(x)} + 1.53y^2z^3sin{(x)}\end{align*}$

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Calculus 3 : Partial Derivatives

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Example Question #1211 : Partial Derivatives

Example Question #1211 : Partial Derivatives

Example Question #1211 : Partial Derivatives

Example Question #1212 : Partial Derivatives

Example Question #1213 : Partial Derivatives

Example Question #1214 : Partial Derivatives

Example Question #1215 : Partial Derivatives

Example Question #1212 : Partial Derivatives

Example Question #152 : Directional Derivatives

Example Question #153 : Directional Derivatives

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